Respuesta :
Answer:
For the function [tex]f(x)=2 x^{2} + 4[/tex] the domain is [tex]\left(-\infty, \infty\right)[/tex] and the range is [tex]\left[4, \infty\right)[/tex].
For the function [tex]g(x)=\sqrt{\frac{x-3}{x-8} }[/tex] the domain is [tex]\left(-\infty, 3\right] \cup \left(8, \infty\right)[/tex] and the range is [tex]\left[0, 1\right) \cup \left(1, \infty\right)[/tex].
Step-by-step explanation:
The domain of a function is the set of input or argument values for which the function is real and defined.
The function [tex]f(x)=2 x^{2} + 4[/tex] has no undefined points nor domain constraints. Therefore, the domain is [tex]\left(-\infty, \infty\right)[/tex].
To find the domain of the function [tex]g(x)=\sqrt{\frac{x-3}{x-8} }[/tex], you must find non-negative values for radicals, this means that we solve
[tex]\frac{x-3}{x-8}\ge \:0\\\\\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{x-3}{x-8}\\\\\left\begin{array}{cccccc}&x<3&x=3&3<x<8&x=8&x>8\\x-3&-&0&+&+&+\\x-8&-&-&-&0&+\\\frac{x-3}{x-8&+&0&-&&+} \end{array}\right\\[/tex]
[tex]\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0\\x<3\quad \mathrm{or}\quad \:x=3\quad \mathrm{or}\quad \:x>8\\\\\mathrm{Merge\:Overlapping\:Intervals}\\x\le \:3\quad \mathrm{or}\quad \:x>8[/tex]
Therefore, the domain is [tex]\left(-\infty, 3\right] \cup \left(8, \infty\right)[/tex]
The range of a function is the set of values of the dependent variable for which a function is defined.
The function [tex]f(x)=2 x^{2} + 4[/tex] has the vertex (0, 4). Therefore, the range is [tex]\left[4, \infty\right)[/tex].
To find the range of the function [tex]g(x)=\sqrt{\frac{x-3}{x-8} }[/tex]. We find the range for the interval [tex]-\infty \:<x\le \:3[/tex] and the range for the interval [tex]8<x<\infty[/tex].
We compute the values of the function at the edges of the interval:
For [tex]\:x=3[/tex], the function value is [tex]f\left(3\right)=0[/tex] and [tex]\lim _{x\to \:-\infty \:}\left(\sqrt{\frac{x-3}{x-8}}\right)=1[/tex].
Therefore, the range of [tex]g(x)=\sqrt{\frac{x-3}{x-8} }[/tex] at the domain interval [tex]-\infty \:<x\le \:3[/tex] is [tex][0,\:1)[/tex]
The [tex]\lim _{x\to \:8+}\left(\sqrt{\frac{x-3}{x-8}}\right)=\infty[/tex] and [tex]\lim _{x\to \infty \:}\left(\sqrt{\frac{x-3}{x-8}}\right)=1[/tex]. Therefore, the range of [tex]g(x)=\sqrt{\frac{x-3}{x-8} }[/tex] at the domain interval [tex]8<x<\infty[/tex] is [tex]\left(1,\:\infty \:\right)[/tex]
Combine the ranges of all domain intervals to obtain the function range: [tex][0,\:1)\cup \left(1,\:\infty \:\right)[/tex]
